if you have 100 liters of $n_2$ and excess $o_2$, how many liters of no can be produced? $n_2 + o_2 -> 2no$…

if you have 100 liters of $n_2$ and excess $o_2$, how many liters of no can be produced? $n_2 + o_2 -> 2no$ 50 liters no 100 liters no 150 liters no 200 liters no

if you have 100 liters of $n_2$ and excess $o_2$, how many liters of no can be produced? $n_2 + o_2 -> 2no$ 50 liters no 100 liters no 150 liters no 200 liters no

Answer

Answer:

D. 200 liters NO

Explanation:

Step1: Analyze the mole - ratio

From the balanced equation $N_2+O_2\rightarrow 2NO$, the mole - ratio of $N_2$ to $NO$ is 1:2.

Step2: Apply Avogadro's law

At the same temperature and pressure, the volume ratio of gases is equal to their mole ratio. Given 100 liters of $N_2$, and since $V(NO)/V(N_2)=n(NO)/n(N_2) = 2/1$.

Step3: Calculate the volume of NO

$V(NO)=2\times V(N_2)$. Substituting $V(N_2) = 100$ liters, we get $V(NO)=2\times100 = 200$ liters.